+5478 x + y = sqrt(x^2+y^2)
x and y could be: 0 and any real number.
If x or y (let's say x) is 0, the equation would be:
0 + y = sqrt(0^2 + y^2)
Which simplifies to:
y = sqrt(y^2)
The sqrt and ^2 cancel out:
y = y
When you have an equation that simplifies to something equals itself, the solution would be all real numbers.
And Rosala, I hope you meant for Anonymous to graph the equation.
If you graph the equation on the calculator, it comes out as a straight line, with the y at 0. This shows that y (or x) is 0 and x (or y) is all real numbers.
+11912 why dont u enter this in the on-site calculator and check for your answer!
+5478 x + y = sqrt(x^2+y^2)
x and y could be: 0 and any real number.
If x or y (let's say x) is 0, the equation would be:
0 + y = sqrt(0^2 + y^2)
Which simplifies to:
y = sqrt(y^2)
The sqrt and ^2 cancel out:
y = y
When you have an equation that simplifies to something equals itself, the solution would be all real numbers.
And Rosala, I hope you meant for Anonymous to graph the equation.
If you graph the equation on the calculator, it comes out as a straight line, with the y at 0. This shows that y (or x) is 0 and x (or y) is all real numbers.
